I never been taught this and it's on the exam papers, how do I do this?
You'll need to get it of the form
$\displaystyle \frac{2(x^2 - 2x - 3) + R}{x^2 - 2x - 3}$ where $\displaystyle R$ is some remainder...
$\displaystyle \frac{2x^2 + 2x + 1}{x^2 - 2x - 3} = \frac{2x^2 - 4x - 6 + 6x + 7}{x^2 - 2x - 3}$
$\displaystyle = \frac{2(x^2 - 2x - 3) + 6x + 7}{x^2 - 2x - 3}$
$\displaystyle = \frac{2(x^2 - 2x - 3)}{x^2 - 2x - 3} + \frac{6x + 7}{x^2 - 2x - 3}$
$\displaystyle = 2 + \frac{6x + 7}{x^2 - 2x - 3}$
$\displaystyle = 2 + \frac{6x + 7}{(x - 3)(x + 1)}$.
Now you need to break down the remainder using the method of partial fractions.
Let $\displaystyle \frac{A}{x - 3} + \frac{B}{x + 1} = \frac{6x + 7}{(x - 3)(x + 1)}$
$\displaystyle \frac{A(x + 1) + B(x - 3)}{(x - 3)(x + 1)} = \frac{6x + 7}{(x - 3)(x + 1)}$
$\displaystyle A(x + 1) + B(x - 3) = 6x + 7$
$\displaystyle Ax + A + Bx - 3B = 6x + 7$
$\displaystyle (A + B)x + A - 3B = 6x + 7$.
So
$\displaystyle A + B = 6$
$\displaystyle A - 3B = 7$.
Subtracting the second equation from the first gives
$\displaystyle 4B = -1$
$\displaystyle B = -\frac{1}{4}$.
Since $\displaystyle A + B = 6$
$\displaystyle A - \frac{1}{4} = 6$
$\displaystyle A = \frac{25}{4}$.
So finally we have:
$\displaystyle \frac{2x^2 + 2x + 1}{x^2 - 2x - 3} = 2 + \frac{6x + 7}{(x - 3)(x + 1)}$
$\displaystyle = 2 + \frac{\frac{25}{4}}{(x - 3)} - \frac{\frac{1}{4}}{(x + 1)}$
$\displaystyle = 2 + \frac{25}{4(x - 3)} - \frac{1}{4(x + 1)}$.
First, divide $\displaystyle x^2-2x-3$ into $\displaystyle 2x^2+2x+1$
You will get a remainder
Hence $\displaystyle \frac{2x^2+2x+1}{x^2-2x-3}=2+\frac{remainder}{(x-3)(x+1)}$
since $\displaystyle x^2-2x-3=(x-3)(x+1)$
Now you need to express the remaining fraction as the sum of partial fractions
$\displaystyle \frac{remainder}{(x-3)(x+1)}=\frac{A}{x-3}+\frac{B}{x+1}$
Solve for A and B.